Doubt on Lax formulation of Korteweg–de Vries equation

Question

The Korteweg–de Vries equation is given by:

$$\frac{\partial u(x,t)}{\partial t}-6u\frac{\partial u(x,t)}{\partial x}+\frac{\partial^3 u(x,t)}{\partial x^3}=0$$

This equation can be formulated using so called Lax pair. Two linear operators, L and M, form a Lax pair if they satisfy:

$$\frac{\partial L}{\partial t}=ML-LM\equiv [M,L]$$

For Korteweg–de Vries equation $L=-\frac{\partial^2}{\partial x^2}+u(x,t)$. To me it seems that when one takes partial derivative of L, it should act on both parts of L. But the result given in the book (Solitons: An Introduction by Drazin and Johnson) is:

$$\frac{\partial L}{\partial t}=\frac{\partial u(x,t)}{\partial t}$$

How is this correct?

Answer 1

I have been able to figure out the answer to my own question. I would post it here in case someone has the same doubt. Suppose L acts on a function $\psi(x,t)$. For the following calculation I would not explicitly write the $x,t$ dependence of $\psi$ and $u$ \begin{align*} L \psi &= -\frac{\partial ^2 \psi}{\partial x^2}+u\psi \\ \frac{\partial}{\partial t}(L \psi)&=-\frac{\partial ^2}{\partial x^2}(\frac{\partial \psi}{\partial t}) + \frac{\partial u}{\partial t}\psi+ u \frac{\partial \psi}{\partial t} \\ L\frac{\partial \psi}{\partial t}+\frac{\partial L}{\partial t}\psi&=(-\frac{\partial ^2}{\partial x^2}+u) (\frac{\partial \psi}{\partial t})+\frac{\partial u}{\partial t}\psi \\ L\frac{\partial \psi}{\partial t}+\frac{\partial L}{\partial t}\psi&= L\frac{\partial \psi}{\partial t}+\frac{\partial u}{\partial t}\psi \\ \frac{\partial L}{\partial t} &= \frac{\partial u}{\partial t} \end{align*}